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2 years ago

Why would my teacher give me a math paper on something I didn't learn yet? Is she okay?

Mathematics

2 answers:

stepan [7]2 years ago

8 0

To see what you know, which is used to find out what you need to learn.

luda_lava [24]2 years ago

5 0

Answer:

your teacher is crazy about she think you are smart

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A truck travels beneath an airplane that is moving 160 km/h at an angle of 54.0 ◦ to the ground. How fast must the truck travel

Maksim231197 [3]

Answer:

2. 94.0 km/h

Step-by-step explanation:

since the airplane is moving at angle of 54° from the ground, the horizontal component of the velocity is given by cos component

speed of air plane= 160 km/hr

horizontal component of velocity v= 160 cos 54°

= 94.045 km/hr

Thus the truck must travel at an speed of 94.0 Km/hr hence second option is correct.

6 0

3 years ago

On a DVD, information is stored on bumps that spiral around the disk. There are 73,000 ridges (with bumps) and 73,000 valleys (w

Juli2301 [7.4K]

Answer:

I pretty sure it's 8cm.

Step-by-step explanation:

I multiplied 73,000 by 0.000074, which resulted in 5.402.
Then, I multiplied 73,000 by 0.000032, which resulted in 2.336.
I added the products together, which gave a sum of 7.738.
Then I rounded to the nearest whole number, which is 8, because the tenth in the decimal (.7) rounds up because it is >4.

5 0

2 years ago

PT=4x+5 and TQ =6x-7

gregori [183]

Answer:

Step-by-step explanation:

6 0

4 years ago

Write a polynomial equation of least degree with roots 2, -1 and 4.

Rasek [7]

A polynomial with roots a and b is (x - a)(x - b).

(x - 2)(x - (-1))(x - 4) = 0

(x - 2)(x + 1)(x - 4) = 0

has roots 2, -1, and 4.

3 0

3 years ago

A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo

djverab [1.8K]

Answer:

a) $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) $31690.7 g/L$

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is $\frac{dy}{dt}=R_{i}-R_{o}$ , where $R_{i}$ is the rate of salt entering and $R_{o}$ is the rate of salt going outside.

Then we have, $R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}$ , and

$R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}$

So we obtain. $\frac{dy}{dt}=4000-\frac{2y}{25}$ , then

$\frac{dy}{dt}+\frac{2y}{25}=4000$ , and using the integrating factor $e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}$ , therefore $(\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}$ , we get $\frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}$ , after integrating both sides $y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C$ , therefore $y(t)= 50000 +Ce^{\frac{-2t}{25}}$ , to find $C$ we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions $y(0)=10$ , so $10= 50000+Ce^{\frac{-0*2}{25}}$

$10=50000+C\\C=10-50000=-49990$

Finally we can write an expression for the amount of salt in the tank at any time t, it is $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) The tank will overflow due Rin>Rout, at a rate of $80 L/min-40L/min=40L/min$ , due we have 500 L to overflow $\frac{500L}{40L/min} =\frac{25}{2} min=t$ , so we can evualuate the expression of a) $y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7$ , is the salt concentration when the tank overflows

4 0

4 years ago